Q. What does the Code mean by an “Effective Ground-Fault Current Path”?
See the answer below.
This answer is based on the 2017 NEC.
A. The effective ground-fault current path is made by bonding together the metal parts of electrical raceways, cables, enclosures, and equipment and to the supply source in a manner that creates a low‑impedance path for ground‑fault current to flow, which facilitates the operation of the circuit overcurrent protection device [Sec. 250.4(A)(5)]. (Fig. 1)
To ensure a low‑impedance ground‑fault current path, all circuit conductors must be grouped together in the same raceway, cable, or trench [300.3(B), 300.5(I) and 300.20(A)].
Because the earth isn’t a low impedance path for fault current, it isn’t suitable to serve as the required effective ground‑fault current path. Therefore, an equipment grounding conductor (EGC) of a type recognized in Sec. 250.118 is required to be installed with all circuits.
Let’s run through an example problem to prove this point.
What’s the maximum fault current that can flow through the earth to the power supply from a 120V ground fault to metal parts of a light pole without an EGC that’s grounded (connected to the earth) via a rod having a contact resistance to the earth of 25 ohms?
Solution:
I = E ÷ R
I = 120V ÷ 25 ohms = 4.80A
DANGER: Because the contact resistance of an electrode to the earth is so high, very little fault current returns to the power supply if the earth is the only fault current return path. Result — the circuit overcurrent protection device won’t open and all metal parts associated with the electrical installation, metal piping, and structural building steel will become and remain energized (Fig. 2).
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